AMC 10 · 2004 · #19

Grade 10 geometry-3d
spatial-visualizationtangent-circlesinradius physical-representationidentify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
A cone has its top cut off by a plane parallel to the base, leaving circular bases of radii 18 and 2. A sphere inside touches both bases and the slanted side. Find the radius of that sphere.

Pick an answer.

(A)
6
(B)
$4\sqrt{5}$
(C)
9
(D)
10
(E)
$6\sqrt{3}$
How to solve
Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships) is the whole game here: slice the solid with a vertical plane through its axis. Because the sphere's center lies on that axis, the slice cuts the sphere in a great circle of the same radius r, and it cuts the solid in an isosceles trapezoid. Every tangency survives the slice, so the 3D question becomes the flat question 'a circle inscribed in a trapezoid'. Tool #1 (Draw a Diagram) then labels that trapezoid with real numbers. Tool #4 (Introduce a Variable) names the missing height h, and Tool #7 (Identify Subproblems) splits the work into two easy pieces: find the slant side from tangent lengths, then find h from a right triangle.

1STEP 1

Slice the solid down its axis

Slicing down the axis turns the solid into a trapezoid with an inscribed circle.

AB = 2 · 18 = 36, DC = 2 · 2 = 4
2STEP 2

Touching both bases sets 2r=h

Touching both parallel sides makes the height exactly two radii.

h = 2r ⟺ r = h/2
3STEP 3

Tangent lengths give the slant side

Equal tangent lengths make the slanted side 20.

BC = BF + FC = BE + CG = 18 + 2 = 20
4STEP 4

Pythagoras on the slanted side

A right triangle then gives the height as 12.

h = √(20² - 16²) = √(144) = 12
5STEP 5

Check the figure is consistent

Opposite side sums agree, so the figure is consistent.

AB + DC = 36+4 = 40 = 20+20 = BC + AD
6STEP 6

Halve the height

Halving the height gives the radius 6, choice (A).

r = h/2 = 12/2 = 6 (A)
Answer
6
The sphere must be wide enough to reach the slanted wall but must still fit inside, so r should sit between the two base radii, and 2 < 6 < 18 holds. A direct coordinate check confirms it: put the bottom base line as the x-axis with the axis of the solid as the y-axis, so B=(18,0), C=(2,12) and the sphere's center is O=(0,6). Line BC is 12x+16y-216=0, and the distance from O to it is (|12 · 0+16 · 6-216|)/(√(12²+16²)) = 120/20 = 6, exactly r, so the circle really is tangent to the slanted side. The larger choices (C) 9, (D) 10, (E) 6√(3)≈ 10.4 would each need a height of 18 or more, which no longer matches a slant side of 20 with a horizontal run of 16.
💡Key takeaway

Slice the solid straight down its axis: the ball becomes a circle squeezed inside a trapezoid, equal tangent pieces make the slanted side 18+2=20, Pythagoras turns that into height 12, and the ball's radius is half of it, 6.

  • Slice the solid down its axis
  • Touching both bases sets 2r=h
  • Tangent lengths give the slant side
  • Pythagoras on the slanted side
  • Check the figure is consistent
  • Halve the height